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1
COLUMNS: BUCKLING
Slide No. 1
Buckling
Introduction– Buckling is a mode of failure generally
resulting from structural instability due to compressive action on the structural member or element involved.
– Examples• Overloaded metal building columns.• Compressive members in bridges.• Roof trusses.• Hull of submarine.
2
Slide No. 2
Buckling
Introduction– Examples (cont’d)
• Metal skin on aircraft fuselages or wings with excessive torsional and/or compressive loading.
• Any thin-walled torque tube.• The thin web of an I-beam with excessive
shear load• A thin flange of an I-beam subjected to
excessive compressive bending effects.
Slide No. 3
Buckling
Introduction– In view of the above-mentioned examples,
it is clear that buckling is a result of compressive action.
– Overall torsion or shear, as was discussed earlier, may cause a localized compressive action that could lead to buckling.
– Examples of buckling for commonly seen and used tools (components) are provided in the next few viewgraphs.
3
Slide No. 4
Buckling
IntroductionFigure 1
Slide No. 5
Buckling
Introduction
Figure 1 (cont’d)
4
Slide No. 6
Buckling
IntroductionFigure 1 (cont’d)
Slide No. 7
Buckling
Introduction– In Fig. 1, (a) to (d) are examples of
temporary or elastic buckling.– While (e) to (h) of the same figure are
examples of plastic buckling– The distinctive feature of buckling is the
catastrophic and often spectacular nature of failure.
5
Slide No. 8
Buckling
IntroductionFigure 2. Reinforced Concrete
Slide No. 9
Buckling
IntroductionFigure 3. Steel Beam Buckling
6
Slide No. 10
Buckling
Introduction– The collapse of a column supporting
stands in a stadium or the roof of a building usually draws large headlines and cries of engineering negligence.
– On a lesser scale, the reader can witness and get a better understanding of buckling by trying to understand a few of the tests shown in Fig. 1.
Slide No. 11
Buckling
The Nature of Buckling– In the previous chapters, we related load to
stress and load to deformation.– For these non-buckling cases of axial,
torsional, bending, and combined loading, the stress or deformation was the significant quantity in failure.
– Buckling of a member is uniquely different in that the quantity significant in failure is
7
Slide No. 12
Buckling
The Nature of Bucklingthe buckling load itself.
– The failure (buckling) load bears no unique relationship to the stress and deformation at failure.
– Our usual approach of deriving a load-stress and load-deformation relations cannot be used here, instead, the approach to find an expression for the buckling load Pcr.
Slide No. 13
Buckling
The Nature of Buckling– Buckling is unique from our other
structural-element considerations in that it results from a state of unstable equilibrium.
– For example, buckling of a long column is not caused by failure of the material of which the column is composed, but by determination of what was a stable state of equilibrium to an unstable one.
8
Slide No. 14
Buckling
The Nature of Buckling– Mechanism of Buckling
• Let’s consider Fig. 4, 5, and 6, and study them very carefully.
• In Fig. 4, some axial load P is applied to the column.
• The column is then given a small deflection by applying the small lateral force F.
• If the load P is sufficiently small, when the force F is removed, the column will go back to its original straight condition.
Slide No. 15
BucklingThe Nature of Buckling– Mechanism of Buckling
crPP < crPP < crPP <
F
BeforeF
DuringF
AfterF
Stable Equilibrium
Figure 4
F
BeforeF
DuringF
AfterF
9
Slide No. 16
Buckling
The Nature of Buckling– Mechanism of Buckling
• The column will go back to its original straight condition just as the ball returns to the bottom of the curved container.
• In Fig. 4 of the ball and the curved container, gravity tends to restore the ball to its original position, while for the column the elasticity of the column itself acts as restoring force.
• This action constitutes stable equilibrium.
LECTURE 26. Columns: Buckling (pinned ends) (10.1 – 10.3) Slide No. 17
Buckling
The Nature of Buckling– Mechanism of Buckling
• The same procedure can be repeated for increased value of the load P until some critical value Pcr is reached, as shown in Fig. 5.
• When the column carries this load, and a lateral force F is applied and removed, the column will remain in the slightly deflected position. The elastic restoring force of the column is not sufficient to return the column to its original
10
Slide No. 18
Buckling
The Nature of Buckling– Mechanism of Buckling
crPP = crPP = crPP =
F
BeforeF
DuringF
AfterF
Precarious Equilibrium
Figure 5
BeforeF
DuringF
AfterF
F
Slide No. 19
Buckling
The Nature of Buckling– Mechanism of Buckling
straight position but is sufficient to prevent excessive deflection of the column.
• In Fig. 5 of the ball and the flat surface, the amount of deflection will depend on the magnitude of the lateral force F.
• Hence, the column can be in equilibrium in an infinite number of slightly bent positions.
• This action constitutes neutral or precarious equilibrium.
11
Slide No. 20
Buckling
The Nature of Buckling– Mechanism of Buckling
• If the column is subjected to an axial compressive load P that exceeds Pcr, as shown in Fig. 6, and a lateral force F is applied and removed, the column will bend considerably.
• That is, the elastic restoring force of the column is not sufficient to prevent a small disturbance from growing into an excessively large deflection.
Slide No. 21
BucklingThe Nature of Buckling– Mechanism of Buckling
crPP > crPP > crPP >
F
BeforeF
DuringF
AfterF
Unstable Equilibrium
Figure 6
BeforeF
DuringF
AfterF
Possiblebuckle
orcollapse
F
Smalldisturbance ν
12
Slide No. 22
Buckling
The Nature of Buckling– Mechanism of Buckling
• Depending on the magnitude of P, the column either will remain in the bent position or will completely collapse and fracture, just as the ball will roll off the curved surface in Fig. 6.
• This type of behavior indicates that for axial loads greater than Pcr, the straight position of a column is one of unstable equilibrium in that a small disturbance will tend to grow into an excessive deformation.
Slide No. 23
Buckling
The Nature of BucklingDefinition“Buckling can be defined as the sudden large deformation of structure due to a slight increase of an existing load under which the structure had exhibited little, if any, deformation before the load was increased.”
13
Slide No. 24
Buckling of Long Straight Columns
Critical Buckling Load– The purpose of this analysis is to
determine the minimum axial compressive load for which a column will experience lateral deflection.
– Governing Differential Equation:• Consider a buckled simply-supported column of
length L under an external axial compression force P, as shown in the left schematic of Fig. 7. The transverse displacement of the buckled column is represented by δ.
Slide No. 25
Buckling of Long Straight Columns
Critical Buckling Load
(a)(b)
PP
Py
Figure 7
y
14
Slide No. 26
Critical Buckling Load– Governing Differential Equation:
• The right schematic of Fig. 7 shows the forces and moments acting on a cross-section in the buckled column. Moment equilibrium on the lower free body yields a solution for the internal bending moment M,
0=+MPy (1)
Buckling of Long Straight Columns
Slide No. 27
Buckling of Long Straight Columns
Critical Buckling Load– Governing Differential Equation (cont’d):
• Recall the relationship between the moment Mand the transverse displacement y for the elastic curve,
• Eliminating M from Eqs. 1 and 2 results in the governing equation for the buckled slender column,
MdxdyEI =2
2(2)
15
Slide No. 28
Buckling of Long Straight Columns
Critical Buckling Load– Governing Differential Equation (cont’d):
– Buckling Solution:• The governing equation is a second order
homogeneous ordinary differential equation with constant coefficients and can be solved by the method of characteristic equations. The solution is found to be,
02
2
=+ yEIP
dxyd
(3)
Slide No. 29
Buckling of Long Straight Columns
Critical Buckling Load– Buckling Solution (cont’d):
• Where p2 = P/EI. The coefficients A and B can be determined by the two boundary conditions, y(0) = 0 and y(L) = 0, which yields,
pxBpxAxy cossin)( += (4)
0sin0
==pLA
B(5)
16
Slide No. 30
Buckling of Long Straight Columns
Critical Buckling Load– Buckling Solution (cont’d):
• The coefficient B is always zero, and for most values of m × L the coefficient A is required to be zero. However, for special cases of m × L, A can be nonzero and the column can be buckled. The restriction on m × L is also a restriction on the values for the loading F; these special values are mathematically called eigenvalues. All other values of F lead to trivial solutions (i.e. zero deformation).
Slide No. 31
Buckling of Long Straight Columns
Critical Buckling Load– Buckling Solution (cont’d):
• Since p2 = P/EI, therefore,Ln
LLLp
npLpL
ππππ
ππππ
,,3 ,2 , ,0
or ,,3 ,2 , ,0
0sin
L
L
=
=⇒=
(6)
( ) ( )2
22
2
22
2
22
2
2
,,3,2, ,0LEIn
LEI
LEI
LEIP ππππ
L= (7)
17
Slide No. 32
Buckling of Long Straight Columns
Critical Buckling Load– Buckling Solution (cont’d):
• Or
• The lowest load that causes buckling is called critical load (n = 1).
L3,2,1,0for 2
=
= nLnEIP π (8)
2
2
LEIPcr
π= (9)
Slide No. 33
Buckling of Long Straight Columns
Critical Buckling Load, PcrThe critical buckling load (Euler Buckling) for a long column is given by
whereE = modulus of elasticity of the materialI = moment of inertia of the cross sectionL = length of column
2
2
LEIPcr
π= (9)
18
Slide No. 34
Buckling of Long Straight Columns
Critical Buckling Load– Equation 9 is usually called Euler's formula.
Although Leonard Euler did publish the governing equation in 1744, J. L. Lagrange is considered the first to show that a non-trivial solution exists only when n is an integer. Thomas Young then suggested the critical load (n = 1) and pointed out the solution was valid when the column is slenderin his 1807 book. The "slender" column idea was not quantitatively developed until A. Considèreperformed a series of 32 tests in 1889.
Slide No. 35
Buckling of Long Straight Columns
Critical Buckling LoadShape function:
• Substituting the expression of P in Eq. 9, into Eq. 4, and noting that B = 0, the shape function for the buckled shape y(x) is mathematically called an eigenfunction, and is given by,
( )
=LxnAxy πsin (10)
19
Slide No. 36
Buckling of Long Straight Columns
Critical Buckling Stress– The critical buckling normal stress σn is
found as follows:When the moment of inertia I in Eq. 9 is replaced by Ar2, the result is
whereA = cross-sectional area of column
r = radius of gyration =
( ) crcr
rLE
AP σπ
== 2
2
/(11)
AI
Slide No. 37
Buckling of Long Straight Columns
Critical Buckling StressThe critical buckling normal stress is given by
Wherer = radius of gyration =
(L/r) = slenderness ratio of column
( )22
/ rLE
crπσ =
AI
(12)
20
Slide No. 38
Buckling of Long Straight Columns
Critical Buckling Load and Stress– The Euler buckling load and stress as
given by Eq. 9 or Eq. 12 agrees well with experiment if the slenderness ratio is large (L/r > 140 for steel columns).
– Short compression members (L/r < 140 for steel columns) can be treated as compression blocks where yielding occurs before buckling.
Slide No. 39
Buckling of Long Straight Columns
Critical Buckling Load and Stress– Many columns lie between these extremes
in which neither solution is applicable.– These intermediate-length columns are
analyzed by using empirical formulas to be described later.
– When calculating the critical buckling for columns, I (or r) should be obtained about the weak axis.
21
Slide No. 40
Buckling of Long Straight Columns
Review of Parallel-Axis Theorem for Radius of Gyration– In dealing with columns that consist of
several rolled standard sections, it is sometimes necessary to compute the radius of gyration for the entire section for the purpose of analyzing the buckling load.
– It was shown that the parallel-axis theorem is a useful tool to calculate the second
Slide No. 41
Buckling of Long Straight Columns
Review of Parallel-Axis Theorem for Radius of Gyration– Moment of area (moment of inertia) about
other axes not passing through the centroid of the overall section.
– In a similar fashion, the parallel-axis theorem can be used to find radii of gyration of a section about different axis not passing through the centroid.
22
Slide No. 42
Buckling of Long Straight Columns
Review of Parallel-Axis Theorem for Radius of Gyration– Consider the two channels, which are
laced a distance of 2a back to back.
C
CC
C xxxx
xxx rAI
AI
AIrII ====⇒=
secsecoverall 22
2
( ) ( ) ( )( ) 22
sec
22sec
overall
22sec
2sec
2sec
2sec
22
222
drA
drAAI
r
drAdArAdAII
C
C
CCC
yyy
y
yyyy
+=+
==⇒
+=+=+=
(13)
(14)
Lacing bars
xx
y
y
2d
xC2a
Slide No. 43
Buckling of Long Straight Columns
Parallel-Axis Theorem for Radius of Gyration
Eqs. 13 and 14 indicate that the radius of gyration for the two channels is the same as that for one channel, and
( )22Cyy xarr
C++= (15)
dxa C =+ where
Lacing bars
xx
y
y
2d
xC2a
23
Slide No. 44
Buckling of Long Straight Columns
Example 1A 3-m column with the cross section shown in Fig. 8 is constructed from two pieces of timber. The timbers are nailed together so that they act as a unit. Determine (a) the slenderness ratio, (b) the Euler buckling load (E = 13 GPa for timber), and (c) the axial stress in the column when Euler load is applied.
Slide No. 45
Buckling of Long Straight Columns
Example 1 (cont’d)
150 mm
150 mm
50 mm 50 mm
50 mm
Figure 8
50 mm
24
Slide No. 46
Buckling of Long Straight Columns
Example 1 (cont’d)Properties of the cross section:
( )( )( ) ( )( )
( )( ) ( )( ) ( )( )
( )( ) ( )( )
3.32000,15
10625.15
mm 10625.151505012150150
121
mm 1013.53251003175150
3112550
31
bottom from mm 0.75000,15
1505075501505025mm 000,15501502
6min
4633
46333
2
=×
===
×=+=
×=−+=
=×++×
=
==
AI
AIr
I
I
y
A
y
x
x
C
150 mm
150 mm
50 mm 50 mm
50 mm
50 mm75 mm
N.A.
Slide No. 47
Buckling of Long Straight Columns
Example 1 (cont’d)(a) Slenderness Ratio:
(b) Euler Buckling Load:
(c) Axial Stress:
9327.32
3000ratio sSlendernes ===rL
( )( )( )
kN 75.2223
10625.15 10132
692
2
2
=××
==−ππ
LEI
P ycr
(C) MPa 85.14101575.222
3 =×== −A
Pcrσ
25
Slide No. 48
Buckling of Long Straight Columns
Example 2A WT6 × 36 structural steel section is used for an 18-ft column. Determine(a) The slenderness ratio.(b) The Euler buckling load. Use E =
29×103 ksi.(c) The axial stress in the column when
Euler load is applied.
Slide No. 49
Buckling of Long Straight Columns
Example 2 (cont’d)For a WT6 × 36 section (see Fig 9, or Appendix B of Textbook:
in 48.1 in 6.10 min2 == rA
( )( )( )
(C) ksi 43.136.104.142 (c)
kips 4.1429.145
6.10000,29/
(b)
(slender) 1469.14548.11218 a)(
2
2
2
===
===
≅=×
=
AP
rLEAP
rL
cr
cr
σ
ππ
26
ENES 220 ©Assakkaf
Buckling of Long Straight Columns
Example 2 (cont’d)
Figure 9
Slide No. 51
Buckling of Long Straight Columns
Example 3Two C229 × 30 structural steel channels are used for a column that is 12 m long. Determine the total compressive load required to buckle the two members if(a) They act independently of each other.
Use E = 200 GPa.(b) They are laced 150 mm back to back
as shown in Fig. 10.
27
Slide No. 52
Buckling of Long Straight Columns
Example 3 (cont’d)
Lacing bars
xx
y
y
150 mm
Figure 10
Slide No. 53
Buckling of Long Straight Columns
Example 3 (cont’d)(a) Two channels act independently:
• If the two channels are not connected and each acts independently, the slenderness ratio is determined by using the minimum radius of gyration rmin of the individual section
• For a C229 × 30 section (see Fig 11, or Appendix B of Textbook):
2min mm 3795 mm 3.16 === Arr y
28
LECTURE 26. Columns: Buckling (pinned ends) (10.1 – 10.3) Slide No. 54ENES 220 ©Assakkaf
Buckling of Long Straight Columns
Example 3 (cont’d)
Figure 11
Slide No. 55
Buckling of Long Straight Columns
Example 3 (cont’d)
• (b) For a C229 × 30 section (see Fig 11, or Appendix B of Textbook):
( )( ) ( ) ( )[ ]
( )kN 27.6N 1064.27
2.736103795 2 10200
/
(slender) 2.7363.16
1012
32
692
2
2
3
=×=××
==
=×
=
−ππrLEAP
rL
cr
mm 1001.1 mm10 3.25
mm 8.14 mm 8.81646
min
×=×=
==
yx
C
IIxr
Lacing bars
xx
y
y
150 mm
xC = 14.8 mm
29
Slide No. 56
Buckling of Long Straight Columns
Example 3 (cont’d)( ) ( )
( ) ( )[ ]
( ) mm 3.9137952
1023.63
mm 1023.638.147537951001.122
7.8137952
106.50mm 106.50103.2522
6
26262
6266
=×
==⇒
×=++×=+=
=×
==⇒×=×==
AI
r
AdII
AIrII
yy
yy
xxxx
C
C
( )( ) ( )[ ]
( )kN 3.694
9.146103795210200
/
9.1467.81
1012 e, therefor,7.81
2
692
2min
2
3
minmin
=××
==∴
=×
===
−ππrLEAP
rLrr
cr
x
Lacing bars
xx
y
y
150 mm
xC = 14.8 mm
Slide No. 57
Buckling of Long Straight Columns
Example 3 (cont’d)– An alternate solution for finding rx and ry:
• Using Eqs. 13 and 15,
• Therefore,
( ) ( ) ( )mm 3.91
8.14753.16
mm 8.812222
=
++=++=
==
Cyy
xx
xarr
rr
C
C
mm 8.81min == xrr
The slight difference in the result is due to round-off errors.
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